Quote:
Originally Posted by JDeee40
could someone help explain the logic in ASM lesson 14 example 14B?
'express trains arrive in a poisson process at a rate of 4 per hour. Local trains arrive in a poisson rate of 12 per hour. Calculate the probability that at least 4 locals will arrive before 3 express'
I understand the probability that a local arrives before express is 4/(4+12) and that the probability of 4 local before 3 express is (6 C 4) (.75^4)(.25^2)
but why does he lock it at 6 trains? ...he says the p of 4 or more of the next 6 trains are local is..
(6 C 4) (.75^4)(.25^2) + (6 C 5) (.75^5)(.25^1) + (6 C 6) (.75^6)
why do we always expect 6? Couldn't there be 5 local and 2 express...6 local and 2 express...7 local and 2 express?

This question was answered at length in another thread.
But in brief, to solve a probability question involving a set of events, you want to break the set up into mutually exclusive subevents, so that the probability of the set is the sum of the probabilities of the subsets.
By looking at all subsets of what the first 6 trains can be, you achieve this goal.
Your method does not lead to mutually exclusive subevents. For example, LLLLLEEL would be both 5 locals and 2 expresses and 6 locals and 2 expresses.